- #1
HeavyMetal
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I have been reading through Mark Srednicki's QFT book because it seems to be well regarded here at Physics Forums. He discusses the Dirac Equation very early on, and then demonstrates that squaring the Hamiltonian will, in fact, return momentum eigenstates in the form of the momentum-energy relation.
He uses the Hamiltonian [itex]H_{ab}=cP_{j}(\alpha^{\ j})_{ab}+mc^2(\beta)_{ab}[/itex]
So it is easy to see that [itex](H_{ab})^2=c^2P_{j}P_{k}(\alpha^{\ j}\alpha^{k})_{ab}+mc^{3}P_{j}(\alpha^{\ j}\beta+\beta\alpha^{\ j})_{ab}+m^{2}c^{4}(\beta^2)_{ab}[/itex]
He then explains that if we choose suitable matrices that satisfy a few equations, that we can obtain the momentum eigenstate in the correct momentum-energy form. These matrices satisfy:
{[itex]\alpha^{\ j},\alpha^{k}[/itex]}[itex]_{ab}=2\delta^{\ jk}\delta_{ab}[/itex]
{[itex]\alpha^{\ j},\beta[/itex]}[itex]_{ab}=0[/itex]
[itex](\beta^2)_{ab}=\delta_{ab}[/itex]
Where brackets represent the anticommutator.
Ultimately, after some arithmetic we find that:
[itex](H^2)_{ab}=\textbf{P}^{2}c^{2}\delta_{ab}+m^{2}c^{4}\delta_{ab}=(\textbf{P}^{2}c^{2}+m^{2}c^{4})\delta_{ab}[/itex]
I'm having trouble seeing the steps that it takes to get there. Substituting in the values obtained from the matrices, I can see that the middle term drops out because the value is zero, i.e. [itex]mc^{3}P_{j}(\alpha^{\ j}\beta+\beta\alpha^{\ j})_{ab}=mc^{3}P_{j}*0=0[/itex].
I can also see that the last term is included because [itex]m^{2}c^{4}(\beta^2)_{ab}=m^{2}c^{4}\delta_{ab}[/itex].
I cannot, however, see why [itex]c^{2}P_{j}P_{k}(\alpha^{\ j}\alpha^{k})_{ab}=\textbf{P}^{2}c^{2}\delta_{ab}[/itex].
I can see that [itex]c^{2}P_{j}P_{k}(\alpha^{\ j}\alpha^{k})_{ab}=c^{2}P_{j}P_{k}\frac{1}{2}[/itex]{[itex]\alpha^{\ j},\alpha^{k}[/itex]}[itex]_{ab}=c^2\textbf{P}^2\delta^{\ jk}\delta_{ab}[/itex]. I am confused as to where the [itex]\delta^{\ jk}[/itex] goes!
Is it simply the fact that [itex]\delta^{\ jk}[/itex] and [itex]\delta_{ab}[/itex] are Kronecker deltas? I reread the previous sections, but I don't think it was mentioned. If this is the case, then I understand where that [itex]\delta^{\ jk}[/itex] goes. More importantly, I tried to work out the three equivalencies above by using Pauli matrices, but I quickly get lost! When {[itex]\alpha^{\ j},\alpha^{k}[/itex]}[itex]_{ab}[/itex] is written, does it refer to a 2 x 2 matrix (a x b), in which each entry is a 2 x 2 Pauli matrix (j x k) and therefore is ultimately a 4 x 4 matrix?
Thanks in advance,
HeavyMetal \m/
He uses the Hamiltonian [itex]H_{ab}=cP_{j}(\alpha^{\ j})_{ab}+mc^2(\beta)_{ab}[/itex]
So it is easy to see that [itex](H_{ab})^2=c^2P_{j}P_{k}(\alpha^{\ j}\alpha^{k})_{ab}+mc^{3}P_{j}(\alpha^{\ j}\beta+\beta\alpha^{\ j})_{ab}+m^{2}c^{4}(\beta^2)_{ab}[/itex]
He then explains that if we choose suitable matrices that satisfy a few equations, that we can obtain the momentum eigenstate in the correct momentum-energy form. These matrices satisfy:
{[itex]\alpha^{\ j},\alpha^{k}[/itex]}[itex]_{ab}=2\delta^{\ jk}\delta_{ab}[/itex]
{[itex]\alpha^{\ j},\beta[/itex]}[itex]_{ab}=0[/itex]
[itex](\beta^2)_{ab}=\delta_{ab}[/itex]
Where brackets represent the anticommutator.
Ultimately, after some arithmetic we find that:
[itex](H^2)_{ab}=\textbf{P}^{2}c^{2}\delta_{ab}+m^{2}c^{4}\delta_{ab}=(\textbf{P}^{2}c^{2}+m^{2}c^{4})\delta_{ab}[/itex]
I'm having trouble seeing the steps that it takes to get there. Substituting in the values obtained from the matrices, I can see that the middle term drops out because the value is zero, i.e. [itex]mc^{3}P_{j}(\alpha^{\ j}\beta+\beta\alpha^{\ j})_{ab}=mc^{3}P_{j}*0=0[/itex].
I can also see that the last term is included because [itex]m^{2}c^{4}(\beta^2)_{ab}=m^{2}c^{4}\delta_{ab}[/itex].
I cannot, however, see why [itex]c^{2}P_{j}P_{k}(\alpha^{\ j}\alpha^{k})_{ab}=\textbf{P}^{2}c^{2}\delta_{ab}[/itex].
I can see that [itex]c^{2}P_{j}P_{k}(\alpha^{\ j}\alpha^{k})_{ab}=c^{2}P_{j}P_{k}\frac{1}{2}[/itex]{[itex]\alpha^{\ j},\alpha^{k}[/itex]}[itex]_{ab}=c^2\textbf{P}^2\delta^{\ jk}\delta_{ab}[/itex]. I am confused as to where the [itex]\delta^{\ jk}[/itex] goes!
Is it simply the fact that [itex]\delta^{\ jk}[/itex] and [itex]\delta_{ab}[/itex] are Kronecker deltas? I reread the previous sections, but I don't think it was mentioned. If this is the case, then I understand where that [itex]\delta^{\ jk}[/itex] goes. More importantly, I tried to work out the three equivalencies above by using Pauli matrices, but I quickly get lost! When {[itex]\alpha^{\ j},\alpha^{k}[/itex]}[itex]_{ab}[/itex] is written, does it refer to a 2 x 2 matrix (a x b), in which each entry is a 2 x 2 Pauli matrix (j x k) and therefore is ultimately a 4 x 4 matrix?
Thanks in advance,
HeavyMetal \m/